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题目链接: http://poj.org/problem?id=3186
Description
FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.
The treats are interesting for many reasons:
- The treats are
numbered 1..N and stored sequentially in single file in a long box that
is open at both ends. On any day, FJ can retrieve one treat from either
end of his stash of treats.
- Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
- The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
- Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.
Given
the values v(i) of each of the treats lined up in order of the index i
in their box, what is the greatest value FJ can receive for them if he
orders their sale optimally?
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
Line 1: The maximum revenue FJ can achieve by selling the treats
Sample Input
5
1
3
1
5
2
Sample Output
43
Hint
Explanation of the sample:
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of
indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
题目并没有看明白，还是看最后的Hint才明白的(雾)，大意是，给一个数列，每次只能从头或从尾取一个数，第i次取的数乘i，求和最大为多少。
状态转移方程 以dp[i][j]表示列头取了i个数，列尾取了j个数，则dp[i][j]只与dp[i - 1][j]和dp[i][j - 1]有关，于是得到状态转移方程dp[i][j] = max( dp[i - 1][j] + val[i - 1] * ( i + j ), dp[i][j - 1] + val[n - j] * ( i + j ) ),此处的val我是按下标从0开始的。
1 #include<iostream>
2 #include<cstring>
3 #include<cmath>
4 #include<algorithm>
5 using namespace std;
6
7 int val[2005];
8 int dp[2005][2005];
9
10 int main(){
11 ios::sync_with_stdio( false );
12
13 int n;
14 while( cin >> n ){
15 for( int i = 0; i < n; i++ )
16 cin >> val[i];
17 memset( dp, 0, sizeof( dp ) );
18
19 dp[1][0] = val[0];
20 for( int i = 2; i <= n; i++ )
21 dp[i][0] = dp[i - 1][0] + val[i - 1] * i;
22
23 dp[0][1] = val[n - 1];
24 for( int i = 2; i <= n; i++ )
25 dp[0][i] = dp[0][i - 1] + val[n - i] * i;
26
27 for( int i = 1; i <= n; i++ )
28 for( int j = 1; j + i <= n; j++ )
29 dp[i][j] = max( dp[i - 1][j] + val[i - 1] * ( i + j ), dp[i][j - 1] + val[n - j] * ( i + j ) );
30
31 int ans = 0;
32 for( int i = 0; i <= n; i++ )
33 ans = max( ans, dp[i][n - i] );
34
35 cout << ans << endl;
36 }
37
38 return 0;
39 }
POJ-3186 Treats for the Cows( DP )
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原文地址：http://www.cnblogs.com/hollowstory/p/5450394.html
